So, I am using https://stackoverflow.com/a/298713/1472828 to put an argument "hands.txt" (my agrv[1], which is a file I wanna open) in my command arguments. I have tried both hands.txt and "hands.txt", neither of them worked.
int FileParsing(vector<Card> & v, char * FileName) {
ifstream ifs;
ifs.open(FileName);
if (!ifs.is_open()){
cout << "file cannot be opened." << endl;
} else {
So I use debugger to step through my main:
int main(int argc, char * argv[]){
if (argc !=2 ){
//ErrorMessage();
} else {
...
Debugger tells me that my argc is 2, which is right, but how come every time the debugger just goes to
cout << "file cannot be opened." << endl;
which means the argument just fails at reading it
ifstream ifs;
ifs.open(FileName);
Is there something I missed or I passed the argument in a wrong way?
p.s. The text file was read perfectly from cmd, so it's not the problem of code.
Got the idea from #WhozCraig, while running your program in cmd, the text file is put under debug directory. But if you run it using debugger, you have to put the text file in the same directory with other cpp and h files.
Related
I'm here to trying to take input from the file. If user run this .exe from cmd and give a filename like example.exe input.txt
then it shows the file and read. But if user doesn't give the file name then it run as simply program's run.
Program is running well when I give the input from cmd during run time it run perfectly, but if I don't give the filename during running this file and run simply example.exe an exception show me the error
exception: invalid null pointer
my code is here:
// inputfile.cpp : Defines the entry point for the console application.
#include "stdafx.h"
#include<iostream>
#include<conio.h>
#include<string>
#include<fstream>
using namespace std;
void main(int argc, char* argv[])
{
try {
if (argc > 0)
{
string filename = argv[1];
ifstream in(filename);
in.open(filename);
if (in.is_open())
{
cout << "file opened, do something with file";
}
else
{
cout << endl << "You have Entered Wrong File Name Or File Not Exist in Project's Library" << endl;
}
}
}
catch (exception e)
{
}
cout << endl << "do with the simple program";
_getch();
}
The logic error is in the line
if (argc > 0)
It needs to be
if (argc > 1)
argv[1] is NULL if the program is invoked without arguments.
argc is at least 1, the first argument being the name of the program. When the program is invoked with one argument, argc is 2 and argv[1] is the first argument.
When there is only one argument(always the exec file itself, e.g. using in the way ./exec_file or just double click the exec file), the argv[1] would throw an exception.
Here are some tips:
Try to learn how to debug the code(IDE like Visual Studio, Ecplise or gdb).
VC++ is not standard c++, which means it only runs on Windows, not cross-platform. You'd better learn to use g++ compiler(linux).
I am new to using the command line to start up a program. I want to call my file, called "file.txt". I've been searching for examples but I have not seen what the command line writes.
int main(int argc, char *argv[])
{
if (argc < 2) {
cerr << "Error: file name argument not given" << endl;
exit(1);
}
ifstream inFile( argv[1], ios::in);
}
Got it. So through the command line I navigate to my .exe and call the file I want after.
C:\Users\me\sampleProject\bin\Debug>test.exe sampleFile.txt
And it works.
Thanks.
I'm trying to open multiple files to compile the data in them. My program compiles but when I run it i get the following error:
terminate called after throwing an instance of 'std::bad_alloc'
what(): std::bad_alloc
Aborted
So far my program is pretty lengthy so I'll just link the parts of it that deal with opening the files.
int main(int argc,char *argv[])
{
vector<Plays> yearsEntered;
Plays *MyPlays = new Plays();
if (argc < 2)
{
cout << "No filenames given." << endl;
return 0;
}
for(int i=1;i < argc; ++i)
{
string filename = argv[i];
cout << filename << endl;
filename.append(".csv");
cout << filename << endl;
ifstream inputFile(filename.c_str(), ios::in);
inputFile.open(filename.c_str());
//Error checking in case file fails to open
if (!inputFile)
{
cout << "Could not open file. " <<
"Try entering another file." << endl;
}
}
I'm not quite sure why I'm getting an error but if I had to guess i'd say it was something to do with the fact that argv[i] is a *char array and I'm setting it equal to a string. Also when i run the program it's run like this: ./Analyze 2009 2010 (etc). When i run it it'll print out the name of the file that I want to open so I know the problem is when it tries to open the file itself. This is my first time asking a question so if there's any convention I failed to follow let me know and I'll try to fix it.
You already opened your files once. You don't need to open them again.
The std::ifstream constructor is opening each file, then your invoking .open() for no reason. Remove the inputFile.open() line.
Change this:
ifstream inputFile(filename.c_str(), ios::in);
inputFile.open(filename.c_str());
To this:
ifstream inputFile(filename.c_str());
Here's my code.
ifstream readFile;
readFile.open("voltages.txt");
if (readFile.fail( ) )
{
cout << "\nCould not open the file ... check to see if voltages.txt exists\n";
system("pause");
return;
}
string tempString = "";
//readFile.seekg(0, ios::beg);
int idx = 0;
if (readFile.is_open())
{
while ( readFile.good() )
{
getline(readFile,tempString);
cout << tempString << endl;
}
readFile.close();
}
If I move the voltages.txt to c:/, then it works fine, but I have it in the same folder as my .exe right now, and when I set a breakpoint at cout << tempString << endl; it shows up just as "". I'd like to keep it in the .exe if possible. As you can see I tried seeking to the beginning, with no luck. Please help this C++ noob! Thank you!
It might be because you are trying to read a local file when the program requires a link to the file from C:/
You can test this by replacing voltages.txt with C:/pathToVoltages/voltages.txt
I would suggest passing the file to your executable by doing this:
int main(int argc, char argv[]){
ifstream readFile;
readFile.open(argv[2]);
/*Rest of the code*/
This assumes the path to the file is given by the first argument you give to the program as in:
./program pathToFile/voltages.txt
Hope this helps
I have a program that creates a file and writes to it using ofstream. I need the program to be able to parse command line parameters later on. But for some reason, it does not create a file when I drag-and-drop a file onto the compiled executable, even if the program doesn't involve any command line parameters at all. If the executable is run normally, it works. So I'm left totally confused. Here is the source:
#include <iostream>
#include <fstream>
using namespace std;
int main ()
{
ofstream outfile;
outfile.open("test.txt");
if(outfile.is_open())
{
outfile << "Test";
outfile.close();
}
else cout << "Unable to open file";
return 0;
}
Does anybody have any ideas? I appreciate any help.
You are not using the command line arguments at all. Recode your main() method to look like this:
int main(int argc, char** argv)
{
if (argc != 2)
{
cout << "Usage: blah.exe file" << endl;
return 1;
}
ofstream outfile;
outfile.open(argv[1]);
if(outfile.is_open())
{
outfile << "Test";
outfile.close();
}
else cout << "Unable to open file";
return 0;
}
Be careful what you drop, your code rewrites the file contents.
The following code does what the OP wants:
#include <iostream>
#include <fstream>
using namespace std;
int main ( int argc, char ** argv )
{
cout << argv[1] << endl;
ofstream outfile;
outfile.open("testzzzzzzz.txt");
if(outfile.is_open())
{
outfile << "Testzzzzz";
outfile.close();
cout << "wrote file"<< endl;
}
else cout << "Unable to open file";
string s;
getline( cin, s );
return 0;
}
It allows drag and drop, but doesn't use the dropped file name in the file open. When you drop a file in it, you get the message
"wrote file"
Unfortunately, at the moment I have no idea where it wrote the file - not in the current directory, definitely. Just going to do a search...
Edit: It creates it in your Documents and Settings directory. So to put it in the current directory, you probably need to explicitly prefix it with "./", but I havent't tested this - I leave it as an exercise for the reader :-)
Since you have not specified a path, the file, test.txt, will be saved to the default path. Just bring up a command prompt (i.e. run cmd.exe) and the command prompt will show you the default path. The file should be in this directory.
You can change the default path by editing the HOMEDRIVE & HOMEPATH environment variables.
Also, you should note the other answers. You should be using argc/argv to specify the output file.
you haven't specified a path for "test.txt" so it will try and create that file in the current working directory of the executable. This will be different when the exe is invoked by dropping a file on it than it is when you run the program normally.
Try giving "test.txt" a full path and see if that works.
edit:
To write your output file to the path that contains the exe, you would use
GetModuleFileName(NULL, ...) to the the full path of the exe,
then PathRemoveFileSpec to strip off the exe name, leaving just the exe path then
PathCombine to append test.txt to the exe path